A certain experimental mathematics program was tried out in 2 classes

We are given that a certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Thus, there were a total of 2 x 32 = 64 classes under this program.

If we let a = the number of teachers teaching one class, b = the number of teachers teaching two classes, and c = the number of teachers teaching 3 classes, we can create the following equations:

a + b + c = 37

a + 2b + 3c = 64

Subtracting equation 1 from equation 2, we have:

(a + 2b + 3c = 64) - (a + b + c = 37)

b + 2c = 27

2c = 27 - b

c = (27 - b)/2

We see that c is the GREATEST when b = 1, and thus (27 - 1)/2 = 26/2 = 13.

We also see that c is the LEAST when b = 27, and thus (27 - 27)/2 = 0/2 = 0.

So, the range of values of n is 0 to 13.

Answer: A

There is also a shortcut in the answers.
One may notice that greatest possible values differ in each answer choice in contrast to the least values, which repeat.
To find out the greatest value you should count the total classes (32*2=64), then subtract the total #of teachers since we know from the question that each teacher taught at least one class (64-37=27). Thus we get a number of the available extra-classes for teachers, and all that we need is just to count how many teachers could take 2 more classes, which is 27/2 = 13.5. So the greatest possible value of the # of teachers who had 3 classes is 13.
Only answer A has this option.

x --> Number of teachers who teach in 3 subjects
y --> Number of teachers who teach in 2 subjects
z --> Number of teachers who teach in 1 subject

Total number of classes = 2 * 32 = 64

x + y + z = 37
3x + 2y + z = 64

Check the possibilities from the answer choices:
Put x = 0 --> y + z = 37 and 2y + z = 64 --> y = 27 and z = 10 (Possible minimum)

Put x = 14 --> y + z = 23 and 2y + z = 22 --> y = -1 (Not possible)

Put x = 13 --> y + z = 24 and 2y + z = 25 --> y = 1 and z = 23 (Possible maximum)

Answer: A

a+2b+3c=64
a+b+c =37

subtracting second equation from first we obtain b+2c = 27
or b = 27 - 2c. This shows c cannot be more than 13, otherwise b would be negative. C can be 0 as well.
Below is a table of values depicting the range.

a---- -b----c
10--- 27---0
23--- -1---13

tuff question:

let's first analyse for min, there are 32*2=64 classes total
there are 37 teachers*2=74. this means that no teacher may had 3 lessons
so 0 is min

lets analyze for max: we have x teachers that toguht 3 lessons 3*x+other teachers*classes=64
other teachers are 37-x
3*x+37-x=64
2x=27
x=13,3
as we look for an integer 13 is the max value

A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8


This is my 2 cents.
I did back-solving and started with B because the q asks for least and greatest.

if n = 14, then 14 teachers cover 42 classes.
Remaining classes are 64-42 = 22
Remaining teachers are 37-14 = 23
As the stem says each teach at least covers 1 teacher, this is not good.

So i moved on to A.

if n=13, then 13 teachers cover 39 classes.
Remaining classes are 64-39 = 25
Remaining teachers are 37-13 = 24, which is good as we have more classes than teachers.
if n=0, then we have 37 teachers cover 64 classes.
As these teachers can teach anywhere between 1-2, we know that we have enough teachers to cover all classes without any teacher teaching 3 classes.

Hence A

Here are a couple of video explanations for this question:

Algebra:



Backsolving:

total class=32*2=64
total teacher=37
least:
distribute 1 class each to 37 teachers. remain=64-37=27
again distribute 1 each class to 27 teachers.
only 27 teachers gets 2 classes. no one gets 3 classes.
least value of p=0
greatest:
distribute 1 class each to 37 teachers. remain=64-37=27
again distribute 2 classes each to 27/2=13 teachers. and 1 class to other teachers.
13 teachers get 3 classes.
max value of p=13
ans: A) 0 and 13

Check the answer choices (ALWAYS check the answer choices before choosing a particular solution strategy)
I see that, for each answer choice, the second value (the greatest value of n) is different. So, let's test some of these values.

Let's start by testing answer choice B (0 and 14)
I'd like to start here, since we're asked to identify the greatest value of n, and answer choice B has the biggest possible value of n.
So, is it possible to have 14 teachers who teach 3 classes?
Well, (14)(3) = 42 classes
There are 64 classes altogether (2 classes in each of the 32 schools, means a total of 64 classes)
So, the number of classes that still require teachers = 64 - 42 = 22

How many teachers are remaining?
So far, 14 of the 37 teachers are accounted for (they're the ones who are teaching 3 classes each)
So, the number of teachers remaining = 37 - 14 = 23
Can these 23 remaining teachers cover the remaining 22 classes?
NO!
Each teacher must teach AT LEAST ONE class. So, there aren't enough classes needed for each teacher to teach at least one class.
So, we can ELIMINATE answer choice B.

IMPORTANT: We were VERY CLOSE with answer choice B. We were just one class short of meeting our goal. So, I am quite confident that the greatest possible values of n is 13 (answer choice A). Let's find out.

We'll test answer choice A (0 and 13)
Well, (13)(3) = 39 classes
There are 64 classes altogether
So, the number of classes that still require teachers = 64 - 39 = 25

So far, 13 of the 37 teachers are accounted for. So, the number of teachers remaining = 37 - 13 = 24
Can these 24 remaining teachers cover the remaining 25 classes?
YES!
23 of the teachers can teach 1 class each, and the other teacher can teach 2 classes.
Since the greatest possible value of n is 13, the correct answer is ...


Cheers,
Brent

Since we have at least 2 classes in each of the 32 schools, this means that there will be a total of \(32 \times 2 = 64\) classes in all.

Given that there are 37 teachers, for the first 37 classes, one unique teacher can be assigned to each of the first batch of classes.

This means that if only one teacher is assigned per class, there will be \(64 – 37 = 27\) classes left. So, some of the 37 teachers will have to teach another class[/i][/color] (and possibly more than one).

It’s possible that 27 of the 37 teachers will each teach an additional class. Thus, the least possible number of teachers who will have to teach 3 classes is zero.

To find the maxim number of teachers that could teach 3 classes, we have to add 2 classes to each teacher who is already teaching 1 class. With 27 classes remaining, we have 13 sets of two additional classes, with one left-over class. Thus, the maximum possible number of teachers who could teach 3 classes is 13. This would leave 1 teacher teaching two classes and 23 teachers teaching one class each.

Therefore, the final answer is .

32 Schools X 2-Classes each = 64 Classes (in total)
Total no. of teachers = 37

Least no. of teachers who teach 3 classes = 0
[since 2*37=74, can cover up 64 classes]

Greatest no. of teachers who teach 3 classes = 13


Why?
No of Teachers who teach 3 classes = 14
Remaining teachers = 37-14 = 23
No of classes covered = 14*3 = 42
No of remaining classes = 22
Hence we will have a case where one teacher will have no class to teach
Hence, its 13 teachers.

An easier way to solve this problem:

1.2 classes in each of 32 schools means total of 2*32=64 classes.
2.Total teachers = 37

Now, MINIMUM value of n(teachers who took three classes)

Suppose all of them 2 classes then 37*2 =74 classes which is more than 64..so we can imagine few taking one classes say 25 people took 2 classes and rest 12 took 1 classes..This means total classe is 64 and value of n=0

Now since we know value of N is 0, eliminate option C,D and E

Now, MAXIMUM value of n(teachers who took three classes)

Focus only on options A and B for which max value is 13 and 14

If you plugin 14 as value of n means 14(teacher)*3=42 classes...classes remaining =64-42=22 and teachers remaning(37-14=23)

So even if one teacher takes one classe we will have an empty class..Hence B is out..

Directly chose A.

Another approach

let x,y & z be the number of teachers who take 1,2 & 3 number of classes respectively. So x+y+z=37 (A) and x+2y+3z=64 (B) (as given in question).
(B)-(A) y+2z=27.
z=(27-y)/2. The minimum value of z can be 0 ie none of the teachers taking three classes. For the maximum value of z, y has to be minimum ie y=0.
if y=0, z=27/2=13.5. Since the number of teachers can be integers only so 13.

So correct answer is A

If this asks for maximum possible value , why does the rounding of take the lower value (13) and not (14) ? Bunuel VeritasKarishma help please

Posted from my mobile device

It has nothing to do with rounding off rules.
If x can be 13.5 at the MOST, can it take the value 14? No.
But if we need the value of x to be an integer, then it means the highest integer value it can take is 13.


Now say there is a completely different question - How many people are needed to complete the work within 3 days?
Say you get x = 13.5. What will be your answer? You cannot have 13.5 people. Will you have 13 people or 14 people? Note that since you need 13.5 people to do the job, 13 will NOT be sufficient to complete it. So you need 14 (even if the job will be completed some time before the end of the 3rd day)

Video solution from Quant Reasoning:
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Answer: Option A

Step-by-Step Video solution by GMATinsight